$f(x) = \dfrac{ \sqrt{ x - 3 } }{ x^2 + 8 x + 12 }$ What is the domain of the real-valued function $f(x)$ ?
Answer: $f(x) = \dfrac{ \sqrt{ x - 3 } }{ x^2 + 8 x + 12 } = \dfrac{ \sqrt{ x - 3 } }{ ( x + 6 )( x + 2 ) }$ First, we need to consider that $f(x)$ is undefined anywhere where the radical is undefined, so the radicand (the expression under the radical) cannot be less than zero. So $x - 3 \geq 0$ , which means $x \geq 3$ Next, we also need to consider that $f(x)$ is undefined anywhere where the denominator is zero. So $x \neq -6$ and $x \neq -2$ However, these last two restrictions are irrelevant since $3 > -6$ and $3 > -2$ and so $x \geq 3$ will ensure that $x \neq -6$ and $x \neq -2$ Combining these restrictions, then, leaves us with simply $x \geq 3$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid x \geq3\, \}$.